4
CHEMICAL ENERGY
4.1 SOME BASIC IDEAS
Energy is stored in chemical substances in a number of different ways. In an atom, energy is stored
inside the nucleus, being the basis for nuclear power. Between atoms, weak and strong bonds are a
store of energy, whether within a molecule or between molecules. Both these forms of energy are
potential energy, since they are capable of release for the performance of work.
Atoms and molecules also possess kinetic energy in the form of movement. They move around
randomly, the freedom and speed depending on the physical state and the temperature. When we heat
a substance, its temperature increases. At the atomic level, this is reflected in an increase in kinetic
energy, but we can only measure the temperature. Within molecules, the bonds that hold the atoms
together are also moving, vibrating, rotating and flexing.
This is an important point: temperature and heat are related, but not the same. The temperature
is the external measure of the internal energy. Heat always flows from a hotter object to a colder one.
Temperature, when discussed in these terms, is best measured in a new scale: degrees Kelvin (K). This
has the same rate of increase as Celsius, but starts at 0K, which is equal to –273.15ºC. This is known
as absolute zero.
When energy is released by a chemical reaction, we are not seeing the creation of new energy,
simply the conversion of energy from one form (chemical potential) into other forms (heat, light,
sound). The First Law of Thermodynamics states that energy cannot be created or destroyed, only
interconverted.
Chemical and physical processes involve the transfer of heat to or form the surroundings. As
you have seen in Chapter 2, heat can be considered as another reactant (where heat is taken in from the
surroundings) – this is known as an endothermic process – or another product (where it is released to
the surroundings) – this is known as an exothermic process.
PRACTICE QUESTION
1.
List all the energy conversions occurring as natural gas is burnt in oxygen.
2.
Can you think of any examples of endothermic and exothermic processes?
Internal energy
Before we go any further, some points should be made clear. You have seen the effect of heat on the
rate of a chemical reaction. Heating up a reaction will increase its rate, regardless of whether it is
endothermic or exothermic. An exothermic reaction gives out heat regardless of whether the
surroundings are hot or cold.
Because exothermic processes naturally produce heat, they will automatically heat themselves
up and cause the rate to increase. Endothermic processes may require heat to keep them moving at a
reasonable rate, because they continue to cool down their surroundings by taking in heat.
The reason that a process releases or takes in heat is due to the difference in internal energy of
the reactants and products, as shown in Figure 4.1.
Reactants
exothermic process
which releases excess
energy from reactants
Products
FIGURE 4.1 Energy states for reactants and products in a process
4. Chemical Energy
In an exothermic process, the products have lower internal energy than the reactants, and the “spare”
energy is released as heat. In an endothermic process, the products have more energy, and energy (as
heat) is taken in during the reaction.
4.2 HEATS OF R EACTION
Different exothermic reactions release different amounts of heat; the same applies for endothermic
processes. The heat change (released or absorbed) is known as the heat of reaction (H) or the
enthalpy change, and is generally measured in kJ/mole of reactant.
Exothermic reactions have negative heats of reaction, while endothermic reactions have positive
values for H. Table 4.1 gives some H values.
TABLE 4.1 Heats of reaction for some processes
Process
H (kJ/mole)
Boiling water at 100ºC
+40.7
Burning methane (town gas)
-890
Burning octane (component of petrol)
-5450
Conversion of iron ore to iron
+232
PRACTICE QUESTION
2.
Which of the reactions in Table 4.1 are endothermic? exothermic?
3.
For the reaction 2Pb(s) + 2O2 (g)  2PbO2 (s), H = -554 kJ
(a) Is this an exothermic or an endothermic reaction?
(b) How much heat is absorbed/evolved when 1 mole of PbO2 is produced from Pb + O2?
4.
How much heat would be released by the combustion of (a) 100 g of methane (FW 16) and 100
g of octane (FW 114)?
5.
What would be the heat of reaction for the condensation of water at 100ºC?
PRACTICAL WORK
To measure the energy produced by the combustion of a series of organic fuels.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Obtain the following equipment: a steel can, a spirit burner and a thermometer.
Weigh the dry and empty can.
Add 100 mL of water to the can, and re-weigh. Measure the temperature of the water.
Set up the can on a tripod, as shown by your teacher.
Weigh the empty spirit burner.
Half fill the burner with methanol, and reweigh. Place the burner under the can.
Light the wick and stir the water with the thermometer. When the temperature has
increased by about 10C, extinguish the flame, measure the temperature accurately
and reweigh the burner.
Replace the water in the can, empty the burner contents into a provided residue bottle,
and dry the wick.
Repeat steps 3-8 with ethanol, 1-propanol and 1-butanol.
Chemistry 2
4.2
4. Chemical Energy
CALCULATIONS
1.
Calculate the mass of water, the temperature rise and mass of fuel used for each test.
2.
Calculate the energy produced in each test using the formula below, and also the
moles of fuel used. The chemical formula for each compound is provided.
Energy (J) = 4.18 * mass of water * temperature rise
Methanol
Ethanol
1-Propanol
1-Butanol
CH3OH
CH3CH2OH
CH3CH2 CH2OH
CH3CH2 CH2 CH2OH
3.
4.
Calculate the heat of combustion for each compound.
Look up the heats of formation for the four fuels and water (as a gas) and carbon
dioxide in the SI Chemical Data Book. Write the balanced equations for each
combustion reaction.
Calculate the theoretical heats of combustion for each
compound, and the percentage error from your measured value to the theoretical
value.
Q1. Suggest causes of differences between your measured and theoretical values for the
heats of combustion.
Q2. Plot a graph of heats of combustion (theoretical) against the number of carbon atoms.
Attach this to your this report. What do you notice? Predict the heat of combustion for
1-hexanol, which has two more carbons than 1-butanol.
Q3. One major cause of the greenhouse effect is carbon dioxide emissions due to
combustion of organic compounds. Which of the four fuels tested is the most
“greenhouse-friendly”?
4.3 HESS’S LAW
If we drive from Newcastle to Sydney directly, then we will obviously use less petrol than if we drive
via Bourke! However, in a chemical reaction, the journey doesn’t effect the amount of energy used or
released as long as the start and end points are the same. A reaction that goes from reactant A to
product B is the same in terms of heat of reaction, regardless of whether you go via an intermediate C
or directly!
To say this a different way, Hess’s Law states that the heat liberated or taken in by a reaction H - is independent of the path taken by the reaction. Furthermore, the value of H for the overall
reaction is equal to the sum of the enthalpy changes for each step. Figure 4.3 attempts to illustrate
why, in the same type of diagram as Figure 4.1.
Energy
D
H2
H3
B
C
A
H1
H4
FIGURE 4.3 Illustration of Hess’s Law
Chemistry 2
4.3
4. Chemical Energy
The overall value for H is the same regardless of whether the reaction proceeds directly from A to B,
or via any number of intermediate steps. In the case of Figure 4.3, H4 = H1 + H2 +H3..
EXAMPLE
Show how Hess’s Law applies to the combustion of carbon to carbon dioxide directly or via
carbon monoxide.
Path A
2C + O2  2CO2
H = -787.4 kJ
Path B
2C + O2  2CO
2CO + O2  2CO2
TOTAL
H = -220.8 kJ
H = -566.6 kJ
H = -787.4 kJ
Hess’s Law can be used to calculate the H for reactions where its measurement is not possible (or
easy), if another route can be found where the H values are known. When you do this, you need to
remember a few things:

get the substances in the data reactions on the same as the reactants and products in the reaction
you are interested in; if you have to reverse a reaction, make sure you change the sign of

if you need to change the number of molecules of a given component, multiply the whole
equation by that factor and also the H value

having done this, add up the equations, cancelling out any substances that occur equally on both
sides of the overall equation

add up the H values
EXAMPLE
Calculate H for the reaction 2S(s) + 3O2 (g)  2SO3(g) Given the following H values:
S(s) + O2 (g)  SO2 (g)
SO3 (g)  SO2
(g)
+ ½O2
(g)
-297 kJ
Eqn 1
+99 kJ
Eqn 2
Step 1: make sure the reactants and products in the equation of interest are on the correct
sides of the data equations.
In this case, equation 2 must be reversed to get sulfur trioxide on the products side
SO2 (g) + ½O2
(g)
 SO3
(g)
-99 kJ (sign changed)
Step 2: Multiply the data equations by a factor to get correct number of each reactant and
product.
In this case, we need two S and two SO 3 so multiply both equations by 2.
2S(s) + 2O2 (g)  2SO2 (g)
-594 kJ
2SO2 (g) + O2 (g)  2SO3 (g)
-198 kJ
Chemistry 2
4.4
4. Chemical Energy
Step 3: Add up the two equations by combining all on the left hand side and all on the right
hand side. Hopefully, any extras cancel out - in this SO2 – and you get the equation you are
after. If not, back to the drawing board!
2S(s) + 2O2 (g) + 2SO2
(g)
+ O2 (g)  2SO3 (g) + 2SO2 (g)
which is the correct equation when you cancel out the 2 SO 2 on each side.
Step 4: Add up the H values to get the value for the overall equation – this is the use of
Hess’s Law.
H = -792 kJ
= -396 kJ/mole S
PRACTICE QUESTION
6.
Given the heats of reaction in Table 4.2, calculate the value of H for the reactions
(a) 2C + H2  C2H2
(b) N2 + 2O2  N2O4
(c) 6C + 3H2  C6H6 given that
TABLE 4.2 Heats of reaction data
Reaction
H (kJ)
2C2H2 + 5O2  4CO2 + 2H2O
-2600
C + O2  CO2
-393.7
2H2 + O2  2H2O
-571.8
½N2 + O2  NO2
+33
N2O4  2NO2
+57
2C6H6 + 15O2  12CO2+ 6H2O
-6550
PRACTICAL WORK
The reaction between hydrochloric acid (aqueous HCl) and solid sodium hydroxide produces
heat. To measure the heat of reaction (i.e. kJ of heat produced per mole of reactant), a
simple system using an insulated reaction vessel (i.e. a styrofoam coffee cup) and a
thermometer is employed.
To confirm Hess's Law, the reaction can also be performed in two steps:
1.
dissolving the NaOH in water
2.
reaction of the aqueous NaOH with the HCl
If Hess's Law is true, then the combined heats of reaction from these two steps should be
equal to that of the overall reaction.
I. General measurements
1.
Record the temperature of a beaker of water that has been standing in the lab. This
will serve as the start temperature.
2.
Pipette 25 mL of the HCl into a pre-weighed conical flask. Record the mass of the
liquid.
Chemistry 2
4.5
4. Chemical Energy
II. Heat of reaction of solid NaOH with aqueous HCl
3.
Weigh out about 1 g of NaOH in pellet form into a styrofoam cup. Record the mass.
This should be done as quickly as possible to avoid uptake of water from the
atmosphere.
4.
Add, by pipette, 25 mL of 1 M HCl. Stir with a 110ºC thermometer, covering the
opening of the cup with plastic film. Note the highest temperature reached by the
solution.
5.
Repeat steps 3 and 4 twice. The cup should be rinsed and reasonably dry before the
next run.
6.
Repeat steps 3-5, using 2 g of NaOH and 2 M HCl instead.
III. Confirmation Of Hess's Law
A. Heat Of Solution
7.
Repeat steps 3-5, using water instead of the HCl.
B. Heat of Reaction Of Solutions
8.
Pipette 25 mL of 1 M NaOH into a styrofoam cup.
9.
Repeat steps 4 and 5.
CALCULATIONS
The following calculations need to be performed for each run in each part.
*
the moles of NaOH involved in the reaction
*
the temperature rise
*
the total heat energy released
*
the heat energy per mole (kJ/mole) of NaOH
For each set of results, calculate the average kJ/mole.
Q1. Compare the heat of reaction results from Pt II (the different amounts of reactants).
How similar are they? Is this to be expected?
Q2. Do the results confirm Hess's Law? Show your working. Suggest reasons for any
errors.
Q3. The theoretical heat of reaction for solutions of HCl and NaOH is -57.4 kJ/mole.
Calculate the relative error in your result.
Q4. Explain how could you determine the heat of reaction between gaseous HCl and solid
NaOH indirectly (using Hess's Law) without actually combining the two species.
4.4 HEATS OF FORMATION
What happens if we don’t know and can’t measure the value of AH for one of the intermediate
reactions in a calculation like those in the previous section. Hess’s Law can’t work if you can’t follow
all the steps. Fortunately, there is an alternative, which uses Hess’s Law differently (and invisibly).
The heat of formation of a compound (Hf) is defined as the change in heat energy (enthalpy)
that occurs when the compound is formed from its elements in their standard state (at 1 atm and 25'C).
For example, the heat of formation of liquid water is –286 kJ per mole of water. This is heat of
reaction when 1 mole of hydrogen gas combines with ½ mole of oxygen gas to form 1 mole of water.
This wouldn’t help us much except, that Hf for a compound is not determined experimentally,
but can be calculated from the energies involved in breaking and forming the various bonds. Hf for
an element in its standard state is deemed to be zero.
To use heats of formation for normal reactions is quite simple: H for a reaction is equal to the
sum of the heats of formation of the products minus the sum of the heats of formation of the reactants.
How Hess’s Law is the basis for this is shown in the example below.
Chemistry 2
4.6
4. Chemical Energy
EXAMPLE
Show how Hess’s Law is used in the heat of formation method for calculating H for the
burning of methane.
Combustion reaction
CH4 + 2O2  CO2 + 2H2O
Eqn 1
Heat of formation equations for all components
C + 2H2  CH4
C + O2  CO2
O2 + H2  H2O
Eqn 2
Eqn 3
Eqn 4
If you reverse equation 2 to get methane on the LHS, and multiply and equation 4 by 2, the
sum of equations 2-4 is equation 1.
EXAMPLE
Calculate the value of H in kJ/mole and kJ/g of methane for the burning of methane using
the heat of formation method, and the data in Table 4.3.
Step 1: Write the balanced equation.
CH4 + 2O2  CO2 + 2H2O (g)
Step 2: Calculate the sum of the heats of formation for the products.
Hf
(products)
= Hf CO2 + 2 Hf H2O (g)
= -394 + 2 x –239
= -872 kJ
Step 3: Calculate the sum of the heats of formation for the reactants.
Hf
(reactants)
= Hf CH4 + 2 Hf O2 (g)
= -68 + 2 x 0
= -68 kJ
Step 4: Calculate the heat of reaction by subtracting step 3 (reactants) from step 2
(products).
H
= -872 – (-68)
= -804 kJ/mole methane (FW 16)
= -50 kJ/g methane
PRACTICE QUESTIONS
7.
The standard enthalpy of formation of PbO (s) is -217kJ. Write a chemical equation for this
process.
8.
Calculate the heat energy required to convert limestone into carbon dioxide and calcium oxide.
9.
What is H for 2AlC13 + 3H2O  A12O3 + 6HCI?
10. What is H for HCl + NaOH NaCl + H20?
11. What is H for 2Fe + 3Cu2+  2Fe3+ + 3Cu?
12. When an organic compound bums in oxygen, it produces carbon dioxide, water and heat.
Calculate the heat released when 10 g of toluene (C 7H8) is burnt.
13. Calculate the heat change when 50 g of silver chloride is precipitated from solution.
14. Compare octane (C8H18) as a fuel to methane on a kJ/g basis.
Chemistry 2
4.7
4. Chemical Energy
TABLE 4.3 Selected heats of formation *
Hf (kJ/mol)
Compound
Ag+
AgCl
AlC13
A12O3
CH4
C7H8
C8H18
CaCO3
CaO
ClCO2
Cu2+
Fe3+
HCl
H2O (g)
H2O (l)
NaCl
NaOH
+106
-127
-704
-1676
-68
+15
-208
-1207
-636
-40
-394
+64
-48
-167
-239
-286
-411
-427
* Hf for other compounds may be found in the CRC Handbook of Physics & Chemistry, but you will to convert
them from kcal/mole to kJ/mole by multiply by 4.18.
4.5 BOND ENERGIES
Standard heats of formation are calculated from the energy changes produced by the forming and
breaking of bonds. The energy required to break or form a particular bond (e.g. C-H) is independent
of the compound. The bond energy is defined as the amount of heat that must be absorbed to break a
bond in a gas phase molecule. Table 4.4 lists the bond energies for some common covalent bonds.
TABLE 4.4 Selected bond energies
Bond
H-H
C-H
C-C
Cl-Cl
I-I
N-H
O-O
O-H
BE (kJ/mol)
436
413
348
243
151
391
139
463
Bond
C=C
N=N
O=O
BE (kJ/mol)
682
418
498
Bond
CC
NN
BE (kJ/mol)
962
941
The greater the bond energy, the more energy require to break the bond, the more stable that bond (and
the molecule containing it) is. Compare stable water (H-O-H) and unstable hydrogen peroxide (H-OO-H). The only difference is the O-O bond in hydrogen peroxide. It has a bond energy of 146
kJ/mole, much less than the 463 kJ/mole for the O-H bond, which is very high, and therefore, hard to
break. For a gas phase reaction, the enthalpy change can be calculated from the sum of the energies of
the bonds broken minus the sum of the energies of the bonds formed.
Chemistry 2
4.8
4. Chemical Energy
EXAMPLE
Calculate Hf for ethane gas, which has the formula CH3-CH3.
The heat of formation equation for ethane is 2C + 3H 2  C2H6
Bonds broken:
Bonds formed:
3 x H-H
6 x C-H
1 x C-C
= 3 x 436 = 1308 kJ
= 6 x 413 = 2478
= 348
Heat of reaction
= bonds broken – bonds formed = 1308 – (2478 + 348)
= -1518 kJ/mole ethane
PRACTICE QUESTIONS
15. List the bonds which are broken, and those which are formed in the gas-phase reactions:
(a) H2 + O2  H2O2
(b) N2 + 3H2  2NH3
16. Calculate the energy of an C=O bond, using the heat of formation of carbon dioxide, which
contains 2 of these bonds only.:
17. Calculate the heats of reactions for reactions in Q15.
4.6 SPONTANEOUS REACTIONS
Not all processes for which equations can be written actually will occur “by themselves”. Only certain
processes are spontaneous: in general, exothermic reactions are more likely to be spontaneous,
because they don’t rely on energy from the surrounding to work.
However, not all exothermic reactions are spontaneous (and not all endothermic reactions are
non-spontaneous) because another factor is involved. This is the so-called entropy of the process,
which is a very complex area, but consider it as a measure of randomness. The Second Law of
Thermodynamics states that in all processes, the tendency is towards an increase in randomness. Work
has to be done to done to prevent this happening.
If a process causes the arrangement of the atoms to be less ordered, then it will help it become a
spontaneous process. A simple example is the endothermic process of ice melting – it is endothermic,
but the ordered crystal structure in the ice breaks down to give the more random liquid state. Entropy
is given the symbol S (for no obvious reason) and its change in a process can be measured.
Whether a process is spontaneous depends, therefore, on these two factors:

heat change – favoured by energy release (negative H)

entropy change – favoured by randomness increase (positive S)
Figure 4.4 summarises these effects on the spontaneity of a process.
H
non-spontaneous
can be either
S
can be either
spontaneous
FIGURE 4.4 Factors influencing spontaneity of a process
Chemistry 2
4.9
4. Chemical Energy
WHAT YOU NEED TO BE ABLE TO DO








define terms associated with chemical energy
perform heat of reaction calculations
explain Hess’s Law
perform Hess’s Law-related calculations
perform heat of formation calculations
explain the significance of different bond energies
perform bond energy calculations
outline the factors associated with spontaneous reactions
Chemistry 2
4.10